Each of the five boys can play against each of the other 4. which would be 20 games,
BUT, that would include the pair of games of
boy1 vs boy2 and boy2 vs boy1, the same game.
So we have to divide by 2 to eliminate all those "doubles"
There will be 10 games

The number is actually low enough for you to list them

1-2
1-3
1-4
1-5

2-3
2-4
2-5

3-4
3-5

4-5 for a total of 10

This is a typical "combination" problem where you are choosing the number of cases of "5 choose 2" or
C(5,2) = 5!/(3!2!) = 10

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